show that $f^{(3)}(c) \ge 3$ for $c\in(-1,1)$ Let $f:I\rightarrow \Bbb{R}$, differetiable three times on the open interval $I$ which contains $[-1,1]$. Also: $f(0) = f(-1) = f'(0) = 0$ and $f(1)=1$.  
Show that there's a point $c \in (-1, 1)$ such that $f^{(3)}(c) \ge 3$
I'd be glad to get a guidace here how to start.
 A: We have from Taylor's Theorem $$f(x) = f(0) + xf'(0) + \frac{x^{2}}{2}f''(0) + \frac{x^{3}}{6}f'''(c)$$ where $c$ is some number between $0$ and $x$ and hence on putting $x = 1$ we get $$1 = f(1) = \frac{f''(0)}{2} + \frac{f'''(c_{1})}{6}\tag {1}$$ where $c_{1} \in (0, 1)$ and putting $x = -1$ we get $$0 = f(-1) = \frac{f''(0)}{2} - \frac{f'''(c_{2})}{6}\tag{2}$$ where $c_{2} \in (-1, 0)$ and subtracting equation $(2)$ from equation $(1)$ we get $$f'''(c_{1}) + f'''(c_{2}) = 6\tag{3}$$ It is now obvious that both $f'''(c_{1})$ and $f'''(c_{2})$ can't be less than $3$ (otherwise their sum will be less than $6$). Hence we must have one of these values greater than or equal to $3$.
A: You can use argument contradiction to show. Suppose $f'''(x)<3$ for $\forall x\in(-1,1)$. In [-1,0], by MVT, we have
$$ f(0)-f(-1)=f'(c_1) $$
where $c_1\in(-1,0)$. So $f(c_1)=0$. Similarly in $[c_1,0]$, we have
$$ f'(0)-f'(c_1)=f''(c_2) $$
where $c_2\in(c_1,0)$. So $f''(c_2)=0$. In [0,1], we have
$$ f(1)-f(0)=f'(b_1) $$
where $b_1\in(0,1)$. So $f(b_1)=1$. In $[0,b_1]$, we have
$$ f'(b_1)-f'(0)=f''(b_2)b_1 $$
where $b_2\in(0,b_1)$. So $f''(b_2)=\frac{1}{b_1}>1$.
Since $f'''(x)$ exists in (0,1), we have that $f''(x)$ is contiuous in $(0,1)$ and hence there exists $a\in[c_2,b_2]$ such that $f'':[c_2,a]\to [0,1]$ is an onto map. Choose $c_2\in(c_1,0)$ such that $c_2$ is the largest. Thus in $ (c_2,a)$, $f''>0$ and hence $f''(x)>0$ in $[0,a]$.
For $x\in[0,1]$, we have
$$ f''(x)-f''(0)=f'''(c_3)(x-0) \tag{1} $$
where $c_3\in(0,x)$. (1) implies $$ f''(x)<3x+1 \tag{2}. $$
Integrating (2) twice from (0,x) and using $f(0)=f'(0)=0$, we have
$$ f(x)<\frac{1}{2}x^3+\frac{1}{2}x^2. \tag{3} $$
Letting $x=1$ in (3), we obtain 
$$ f(1)<\frac{1}{2}+\frac{1}{2}=1$$
which is against $f(1)=1$.
A: This is only a partial answer. I will try to prove this for a polynomial function $f(x)$. The degree of this polynomial is at least 3. Zero is a double root and -1 is a single root:
$$f(x)=c(x)x^2(x+1)$$
The first thing I want to prove is that there exists an $a<0$ with $f''(a)=0$ and a $b>0$ with $f''(b)>1$. Remember that $f^{(3)}$ exists, so $f''(x)$ is continuous. As a result of continuity, $f''(x)$ reaches all the values between $0$ and $1$ in the interval $(a,b)$. Choose the largest such $a$ and the smallest such $b$, so that $f''(x)$ reaches in $(a,b)$ every value $y \in (0,1)$ exactly once. The image of $f''$ for $x \in (a,b)$ is thus $(0,1)$. 
(1) Proof that there is an $a \in (-1,0)$ with $f''(a)=0$: 
$f(-1)=0$ and $f(0)=0$ so by Rolle's theorem there is a $c \in (-1,0)$ with $f'(c)=0$.
$f'(c)=0$ and $f'(0)=0$ so by Rolle's theorem there is an $a \in (c,0)$ with $f''(a)=0$.
(2) Proof that there is a $b \in (0,1)$ with $f''(b)>1$:
$f(0)=0$ and $f(1)=1$ so by Lagrange's theorem there is a $d \in (0,1)$ with $f'(d)=1$.
$f'(0)=0$ and $f'(d)=1$ so by Lagrange's theorem there is an $b \in (0,d)$ with $1=f''(b)d$. Since $0<d<1$, $f''(b)>1$. 
Equipped with this information, I want to prove that there exists an $x \in (0,b)$ with $f^{(3)}(x)>3$:
$$f^{(3)}(x)=c'''(x)(x^3+x)+3c''(x)(3x^2+1)+3c'(x)(6x)+6c(x)$$
Remember that 


*

*$\forall x \in (0,b)$: $f(x) > 0$ and the same must be true for $c(x)$. You can proof this by contradiction. 

*$\forall x \in (0,b)$: $f'(x) > 0$ and the same must be true for $c'(x)$ (not completely sure how to proof this).

*As we already know: $\forall x \in (0,b)$: $f''(x) > 0$ and the same must be true for $c''(x)$.

*$\forall x \in (0,b)$: $f^{(3)}(x) \geq 0$ (because $f''(x)$ is monotone increasing in $(a,b)$) and the same must be true for $c''(x)$.


If those statements are true, then $\forall x \in (0,b): f^{(3)}(x)>3(3x^2+1)>3$.
I know this is not an incomplete proof, but nevertheless I hope it helps you. 
